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find the area of the figure bonded by $y=x^2-2x+5$ , $y=5x-5$

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Domain and range of quadratic functions: equations

Full solution

Q. find the area of the figure bonded by

y=x^2-2x+5

,

y=5x-5

Set Equations Equal: Find the points of intersection between $y=x^2-2x+5$ and $y=5x-5$ . Set the equations equal to each other: $x^2-2x+5=5x-5$ .
Rearrange and Simplify: Rearrange the equation to find the $x$ -values of the intersection points: $x^2-2x+5-(5x-5)=0$ . Simplify: $x^2-7x+10=0$ .
Factor Quadratic Equation: Factor the quadratic equation: $(x-5)(x-2)=0$ . Solve for $x$ : $x=5$ or $x=2$ .
Solve for x: Calculate the definite integral of the top function minus the bottom function from the left intersection point to the right intersection point. $\newline$ Integrate from $x=2$ to $x=5$ : $\int_{2}^{5} (5x-5)-(x^2-2x+5) \, dx$ .
Integrate Functions: Simplify the integrand: $\int(5x-5-x^2+2x-5) \, dx$ . Combine like terms: $\int(-x^2+7x-10) \, dx$ .
Simplify Integrands: Find the antiderivative: $-\frac{x^3}{3} + \frac{7x^2}{2} - 10x$ .
Find Antiderivative: Evaluate the antiderivative from $x=2$ to $x=5$ . Plug in the limits of integration: $[(-5^3/3 + 7\cdot5^2/2 - 10\cdot5) - (-2^3/3 + 7\cdot2^2/2 - 10\cdot2)]$ .
Evaluate Antiderivative: Calculate the values: $[\left(-\frac{125}{3} + \frac{175}{2} - 50\right) - \left(-\frac{8}{3} + \frac{28}{2} - 20\right)]$ . $\newline$ Simplify the expression.

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